MySQL Query gone crazy

beettlle

I am running a query to a MySQL db and am getting a non_NULL value in my query variable which means it is supposed to contain the results. When I run the mysql_fetch_array function in it I get nothing out of it. I don't understand what I'm doing wrong.
Source code below.(yes I am using the FULLTEXT index)

/ file with info on connection; "$dbh" is the variable with the connection info/
include("connect.inc") ;

$results = mysql_query("SELECT id FROM table WHERE MATCH (head,body) AGAINST ('php')", $dbh) ;

//$results != NULL when I evaluate it here

$row = mysql_fetch_array($results) ;

printf("(%s)", $row[id]) ;

//output of printf = ()

junebug

put them through a loop like so:

while($row = mysql_fetch_array($results))
{
echo $row['id'];
}

[deleted]

You are making a tiny mistake.

mysql_query() only returns TRUE/FALSE if the query worked/didn't work.

This says NOTHING about how many results the query returned.

Check out mysql_num_rows() in the manual.

beettlle

Been there done that. I still don't get any results. Anyway the DB has only one entry(test table). Any other ideas that might help?

beettlle

The relevant information I want out of the db is the content of the rows from the query and the MATCH statement. The NULL value returned my the mysql_query on to the $results variable tells me that there were no matches. If there are any matches then the variable $results would have a value other then NUL. Or at least that's my understanding, am I right?
Thanks for the help, =)

[deleted]

sigh.
What did I just tell you?

mysql_query does NOT return ANY information about how many records were found.

It only returns TRUE if the query worked, and FALSE if the query did not work ie: you made a mistake in the SQL syntax.

Read the PHP manual about the mysql functions.

beettlle

OK, I know that and that's what I tried to explain in the last post.(not a native english speaker) I found out tryng this.

$result = mysql_query("SELECT * FROM fake_table", $dbh) ;

if ($results == NULL) {
	echo "NULL value" ;
} elseif ($results == "NULL") {
	echo "\"NULL\" text" ;
} elseif ($results == 0) {
	echo "0" ;
} elseif ($results == 1) {
	echo "1" ;
} else {
	echo "no idea: $result" ;
}

Becasue the db table is fake(a table that I havn't created in the db) there is no way the $result will contain a real query and I got "NULL value" from the if statement.

When I tried:

$result = mysql_query("SELECT * from real_table WHEN id = 2000", $dbh) ;

if ($results == NULL) {
	echo "NULL value" ;
} elseif ($results == "NULL") {
	echo "\"NULL\" text" ;
} elseif ($results == 0) {
	echo "0" ;
} elseif ($results == 1) {
	echo "1" ;
} else {
	echo "no idea: $result" ;
}

Becaue the table is a real table(one that I created and know has 2 entries) and I know the id comparison is not going to give me anything back because there are no id values = 2000, I should get a "NULL value" back from the if statemnt and that's exactly what I get.

Now the question is: what happens when I don't get a NULL for the mysql_query function ($results != NULL) but when I run the mysql_fetch_array function I get nothing? I have tried using the mysql_fetch_array and mysql_fetch_row to get the info out of $results but I get nothing. I even got PHP to print out the query it's using to the screen, then copy/paste-ed it over to a straight, command prompt, mysql session to see if the query was wrong(syntax and such). But from the command line I get the results I expect. The query does show me that the word "PHP is in the first row and not on the second. So I realy don't know what's going on.

beettlle

Thanks a lot for the help. The problem turned out to be the MySQL DB instead of the PHP code. I dumped the table with mysqladmin, deleted teh table, then created it again and then jsut put all the infomation back into it. Now it works like a charm. I wonder what it was . . .